Twisted de-Rham cohomology and Eilenberg-Mac Lane spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:51:43Z http://mathoverflow.net/feeds/question/82376 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82376/twisted-de-rham-cohomology-and-eilenberg-mac-lane-spaces Twisted de-Rham cohomology and Eilenberg-Mac Lane spaces Fabio 2011-12-01T12:45:20Z 2011-12-10T10:17:52Z <p>One can define twisted cohomology theories via bundles of classifying spaces. In particular, given a cohomology theory $h^{*}$ and a corresponding $\Omega$-spectrum $E_{n}, \varepsilon_{n}$, we can consider on a space $X$ a bundle with fiber $E_{n}$, and define a twisted version of $h^{n}$ as the set of homotopy classes of sections of such a bundle. If the bundle is trivial, we recover the ordinary cohomology theory.</p> <p>Let us consider a manifold $X$ and its de-Rham cohomology. I suppose there are no problems in defining the Eilenberg-MacLane spaces $K(\mathbb{R}, n)$, even if $\mathbb{R}$ is not countable. For a fixed odd-dimensional form $H$, we can define the (even and odd) twisted de-Rham cohomology groups, via the twisted coboundary $d + H\wedge$, for $d$ the de-Rham differential.</p> <p>Is there a way to relate the two previous approaches? In particular, given an odd-form $H$, is there a suitable bundle of real Eilenberg-MacLane spaces, whose homotopy classes of sections correspond to the twisted cohomology classes defined via $d + H\wedge$?</p> http://mathoverflow.net/questions/82376/twisted-de-rham-cohomology-and-eilenberg-mac-lane-spaces/82394#82394 Answer by Fernando Muro for Twisted de-Rham cohomology and Eilenberg-Mac Lane spaces Fernando Muro 2011-12-01T16:39:35Z 2011-12-01T18:02:30Z <p>You cannot compare both cohomologies unless $H$ has degree $1$, because otherwise the cohomology of $d+H\wedge$ is not $\mathbb{Z}$-graded.</p> <p>If $H$ has degree $1$ then the cohomology of $d+H\wedge$ is the cohomology of $X$ with local coefficients corresponding to the flat line bundle with $1$-form $H$.</p> <p>The only twisted coefficients associated to singular cohomology with real coefficients are the usual local coefficiens. This follows from the fact that, for $E_n=K(\mathbb{R},n)$, the bundles over $X$ with fiber $E_n$ are classified by $B(\operatorname{Aut}^h(E_n))=K(\operatorname{Aut}\mathbb{R},1)$. Here $\operatorname{Aut}^h$ is the topological group (or $A$-infinity space) of self-homotopy equivalences, and $\operatorname{Aut}$ is just an automorphism group. Therefore, bundles over $X$ with fiber $E_n$ are classified by homomorphisms $\pi_1(X)\rightarrow \operatorname{Aut}(\mathbb{R})$, which correspond to local systems.</p> http://mathoverflow.net/questions/82376/twisted-de-rham-cohomology-and-eilenberg-mac-lane-spaces/82435#82435 Answer by ubunke for Twisted de-Rham cohomology and Eilenberg-Mac Lane spaces ubunke 2011-12-02T07:39:44Z 2011-12-10T10:17:52Z <p>You should turn your d+H into a flat superconnection of degree one. Here is the example which works for K-theory. You consider the $\mathbb{Z}$-graded complex $\Omega(M)[b,b^{-1}]$ with $b$ of degree $-2$ and define the superconnection by $d+bH$ for the closed $3$-form $H$. There is an equivalence between the $\infty$-categories of such superconnections and bundles of chain complexes (representations of the singular complex of your underlying manifold), see <a href="http://arxiv.org/abs/0908.2843" rel="nofollow">Block-Smith</a>. If you then apply the Eilenberg-MacLane equivalence between the categories of chain complexes and $H\mathbb{Z}$-modules, then you get a bundle of $H\mathbb{Z}$-modules. This (or the bundle of its $\infty$-loop spaces) is what you are lokking for. Actually, all these steps are equivalences so that you can go backwards.</p>